Giữa 2 bản kim loại đặt song song nằm ngang, tích điện trái dấu, có địên thế U1=1000V , d=1cm .Ở đúng giữa khoảng cÁch 2 bản có 1 giọt thủy ngân nằm lơ lửng. Hiệu điện thế giảm còn U2=995V. Hỏi sau bao lÂu giọt thủy ngân rơi xuống bản dươis

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Đáp án: \(t=0,45s\)

Giải thích các bước giải:

Ta có:

+ Khi giọt thủy ngân cân bằng: \(P = {F_1} \Leftrightarrow mg = q{E_1} = q\dfrac{{{U_1}}}{d}\)

\( \Rightarrow m = q\dfrac{{{U_1}}}{{gd}}\) (1)

+ Khi hiệu điện thế giảm còn \({U_2} = 995V\)

\( \Rightarrow \) Giọt thủy ngân rơi Theo định luật II – Newton: \(P - {F_2} = ma\)

\( \Rightarrow a = \dfrac{{P - {F_2}}}{m} = \dfrac{{mg - q{E_2}}}{m} = g - q\dfrac{{{U_2}}}{{md}}\) (2)

+ Thế (1) vào (2) ta được:

\(a = g - q\dfrac{{{U_2}}}{{q\dfrac{{{U_1}}}{{gd}}.d}} = g - g\dfrac{{{U_2}}}{{{U_1}}}\)

+ Phương trình quãng đường giọt thủy ngân khi đó:

\(s = \dfrac{1}{2}a{t^2}\)

\(\begin{array}{l} \Leftrightarrow \dfrac{d}{2} = \dfrac{1}{2}a{t^2}\\ \Rightarrow t = \sqrt {\dfrac{d}{a}} = \sqrt {\dfrac{d}{{g - g\dfrac{{{U_2}}}{{{U_1}}}}}} \\ = \sqrt {\dfrac{{0,01}}{{10 - 10\dfrac{{995}}{{1000}}}}} \approx 0,45s\end{array}\)

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